If three cells are connected in parallel as shown in figure, then equivalent emf would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( E_{eq} = \displaystyle\sum_{i=1}^{3} E_i/r_i \Big/ \sum_{i=1}^{3} 1/r_i \)
The three cells (emf \( E_i \), internal resistance \( r_i \)) are joined in parallel, so they share the same terminal voltage \( V \) across the external resistor R. Apply Kirchhoff's junction rule to the currents of the cells.
The current given by cell \( i \) is
\( \displaystyle I_i = \frac{E_i - V}{r_i} \)
Adding the three currents gives the current \( I \) through R:
\( \displaystyle I = \sum_{i=1}^{3}\frac{E_i}{r_i} - V\sum_{i=1}^{3}\frac{1}{r_i} \)
Rearranging into the form \( V = E_{eq} - I\,r_{eq} \) of a single cell:
\( \displaystyle V = \frac{\sum E_i/r_i}{\sum 1/r_i} - I\,\frac{1}{\sum 1/r_i} \)
So the combination acts as one cell with
\( \displaystyle E_{eq} = \frac{\sum_{i=1}^{3} E_i/r_i}{\sum_{i=1}^{3} 1/r_i} \qquad \frac{1}{r_{eq}} = \sum_{i=1}^{3}\frac{1}{r_i} \)
Shortcut: Check with three identical cells (\( E_i = E \), \( r_i = r \)): cells in parallel must give \( E_{eq} = E \). Option (3) gives \( (3E/r)/(3/r) = E \). Option (1) gives \( E/r^{2} \), Option (2) gives \( r \) and Option (4) gives \( 1/E \); none of these even has the unit of emf.
NCERT (2026-27):
Cells – “… Equations (3.56) and (3.57) can be extended easily. If there are n cells of emf ε1, . . . εn and of internal resistances r1,.... rn respectively, connected in parallel, the combination is equivalent to a single cell of emf εeq and internal resistance req, such that
\( \displaystyle \frac{1}{r_{eq}} = \frac{1}{r_1} + \ldots + \frac{1}{r_n} \qquad (3.58) \)
\( \displaystyle \frac{\varepsilon_{eq}}{r_{eq}} = \frac{\varepsilon_1}{r_1} + \ldots + \frac{\varepsilon_n}{r_n} \qquad (3.59) \) ”